Why is internal resistance of battery considered outside the terminals although it is present between the terminals inside the battery In ideal battery the internal resistance is zero whereas in non-ideal battery there is some internal resistance now this internal resistance is due to the battery material (electrolyte) and is present inside the battery between the terminals then why do we represent and eventually do calculations by considering that internal resistance to be connected with battery terminals externally. I’m totally unable to get the point. Please help
 A: Suppose that you have a real battery and set up the left hand circuit.

By altering the value of the resistance in the circuit readings of the voltage actross the terminals of the battery $V$ and the current passing through the battery $I$ can be made.
The resulting graph will be approximately a straight line of intercept on the voltage axis of $\mathcal E$ and gradient $-r$.
So to make life easier a model of the real battery consists of an ideal battery of voltage $\mathcal E$ which does not depend on the current passing through it and a resistor $r$ in series with it often called the internal or source resistance.
If that model replaces the real battery in the left hand circuit the voltage against current characteristic of the model battery would be exactly the same as that of the real battery.
Such a model will simplify circuit calculations but in reality there is no ideal battery with a series resistor inside a real battery!
A: For learning about internal resistance we often draw circuits with the (two-stroke) battery symbol next to a resistor representing the internal resistance – and in series with it and the rest of the circuit. In such diagrams the battery symbol does not represent the whole battery! It represents just the emf-giving part of the battery. It would be sensible to put a dotted box around the battery symbol and the internal resistance 'resistor' to indicate that they are both inside the battery.
Of course the emf of the battery cannot in practice be separated from the battery's emf: we can't stick the probes of a multimeter into a battery and measure either its emf or its internal resistance separately. But the circuit does behave as if there is a separate source of emf and internal resistance in series with each other.
A: It's just 'empirical' (a rule which is good enough to predict observations, but is not necessarily related to the actual cause). Observations of the current driven by a battery and its voltage show a roughly linear relationship which can't be explained by the external resistance in the circuit alone, but are consistent with an 'ideal' battery plus an additional resistance. Although the chemical processes which cause this (now labelled) 'internal resistance' are entirely different from those causing resistance in the external circuit, a linear approximation works well well enough and explains what is seen. Once we decide to use this as an assumption, then we can use 'ohms law' and draw the circuits as  an 'ideal battery' plus an 'internal resistance' and get useable results. But it is simply a convention which gets good enough answers, and shouldn't be confused with 'where the resistance sits' in reality.
A: Replacing a real battery by an ideal battery (voltage $V_0$)
and an internal resistance ($R$) is just a model.
You can apply Kirchhoff's voltage law to this model and derive
a relation  between the real voltage $V$ and current $I$:
$$V=V_0 - R I. \tag{1}$$
On the other hand:
You can take a real battery, and measure
the real voltage $V$ and current $I$ for various
external loads connected to the battery.
You may get a $V$-$I$ plot like this:

(image from djb microtech ltd - internal resistance)
This plot is equivalent to the formula (1) above,
when you take $V_0=1.53$ V and $R=0.421$ Ohm.
So the ultimate justification for the simple theoretical model
(an ideal battery and internal resistance) is
that it matches the experimental observations
(measured voltage and current).
Of course, this doesn't mean that the ideal voltage $V_0$
and the internal resistance $R$ are separate entities inside a real battery.
From electrochemistry of galvanic cells we are aware these two things
are not separate. Both are caused by complex processes on the
electrodes and in the electrolyt of the real battery.
But for predicting the measured $V$-$I$ relation we don't
need to care about these details.
A: 
Why does one represent the internal resistance of a non-ideal battery as external?

It is just a way of representing a non-ideal battery. Wherever the battery goes, the resistance goes with it, then why worry if one is showing it inside or outside. It will be considered in the Kirchhoff's voltage law anyhow.
An uncanny failure of this representation would be applying loop law to a circuit consisting of a resistor between the electrodes of a battery because the law wouldn't include the internal resistance (that is shown as outside the terminals) then. This can be tackled with extra care to consider the internal resistance in the law expression or simply changing the representation.
A: Because there is no potential difference inside of the cell due to the fact that the battery does not form a closed circuit. The circuit is internally separated, or 'terminated', at the posts. Voltage is defined as the difference between the electric potential at two points, or the work required per unit of charge to move one unit of charge between those points; connect the battery terminals and you get electron flow across the connecting structure, which is limited by the resistance of the reactive components in the circuit. The connecting mechanism must be included in the calculation, due to the fact that electromotive conduction is a diabatic(thermally inclusive) process.
